3a)Given the following bond enthalpies (bond energies) in kJ/mol C-H 412; C-C 347; O-H 464; O=O 498; C=O 805; C-O358 Enthalpy of combustion = ?
C2H6 + 7/2 O2 → 2CO2 +3H2O
Finding ∆Hcombustion= ?
∆Hcombustion
∑Bonds energy of reactants - ∑Bonds energy of products
[6 (C-C) + 1(C-C) + 7/2 (OO) ] – [2(2(C=O)) + 3(2(O-H))]
=[6(412kJ/mol) + 1 (347 kJ/mol) + 7/2 (496kJ/mol)] – [4(805kJ/mol) + 6(464kJ/mol)]
= 4562kJ/mol – 6004kJ/mol
= -1442kJ/mol
Therefore, the enthalpy of combustion of ethane is -1442kJ/mol.
4)Given the following standard enthalpies of combustion C (s) -393kJ/mol; H2(g) -285.6kJ/mol; C2H6 (g) -1560kJ/mol Calculate the standard enthalpies of formation of ethane.
1) C(s) +O2 (g) → CO2 (g) ∆Hcomb = -393kJ
2) H2 (g) + ½ O2 (g)→ H2O (g) ∆Hcomb = -285.6kJ
3) C2H6 (g) + 7/2 O2 (g) → 2CO2 (g) + 3H2O (l) ∆Hcomb = -1560kJ
Formation equation for C2H6 (g)
2C (s) + 3H2(g) → C2H6 (g) ∆Hformation = ?
Using Hess’s Law:
a) multiply 1) by 2
b) multiply 2) by 3
c) multiply 3) by -1 (reverse the equation)
1) 2C (s) + 2O2 (g) → 2CO2 (g) ∆Hcomb = -786kJ
2) 3H2 (g) + 3/2 O2 (g) → 3H2O (g) ∆Hcomb = -856.8kJ
3) 2CO2 (g) + 3H2O → C2H6 (g) + 7/2 O2 (g) ∆Hcomb = 1560kJ
After cancelling out reactants and products that are same, we get:
2C (s) + 3H2(g) → C2H6 (g) ∆Hformation =-786kJ -856.8kJ +1560kJ -82.8kJ
2C (s) + 3H2(g) → C2H6 (g) ∆Hformation = -82.8kJ
Therefore, the standard enthalpy of formation is -82.8kJ.
7) The Activation energy from A to B is:
350kJ - 250kJ
=100kJ
Therefore, the activation energy for reaction A to B is 100kJ
3a) Given the following bond enthalpies (bond energies) in kJ/mol
C-H 412; C-C 347; O-H 464; O=O 498; C=O 805; C-O358
Enthalpy of combustion = ?
C2H6 + 7/2 O2 → 2CO2 +3H2O
Finding ∆Hcombustion= ?
∆Hcombustion
∑Bonds energy of reactants - ∑Bonds energy of products
[6 (C-C) + 1(C-C) + 7/2 (OO) ] – [2(2(C=O)) + 3(2(O-H))]=[6(412kJ/mol) + 1 (347 kJ/mol) + 7/2 (496kJ/mol)] – [4(805kJ/mol) + 6(464kJ/mol)]
= 4562kJ/mol – 6004kJ/mol
= -1442kJ/mol
Therefore, the enthalpy of combustion of ethane is -1442kJ/mol.
4)Given the following standard enthalpies of combustion
C (s) -393kJ/mol; H2(g) -285.6kJ/mol; C2H6 (g) -1560kJ/mol
Calculate the standard enthalpies of formation of ethane.
1) C(s) +O2 (g) → CO2 (g) ∆Hcomb = -393kJ
2) H2 (g) + ½ O2 (g)→ H2O (g) ∆Hcomb = -285.6kJ
3) C2H6 (g) + 7/2 O2 (g) → 2CO2 (g) + 3H2O (l) ∆Hcomb = -1560kJ
Formation equation for C2H6 (g)
2C (s) + 3H2(g) → C2H6 (g) ∆Hformation = ?
Using Hess’s Law:
a) multiply 1) by 2
b) multiply 2) by 3
c) multiply 3) by -1 (reverse the equation)
1) 2C (s) + 2O2 (g) → 2CO2 (g) ∆Hcomb = -786kJ
2) 3H2 (g) + 3/2 O2 (g) → 3H2O (g) ∆Hcomb = -856.8kJ
3) 2CO2 (g) + 3H2O → C2H6 (g) + 7/2 O2 (g) ∆Hcomb = 1560kJ
After cancelling out reactants and products that are same, we get:
2C (s) + 3H2(g) → C2H6 (g) ∆Hformation =-786kJ -856.8kJ +1560kJ -82.8kJ
2C (s) + 3H2(g) → C2H6 (g) ∆Hformation = -82.8kJ
Therefore, the standard enthalpy of formation is -82.8kJ.
7) The Activation energy from A to B is:
350kJ - 250kJ
=100kJ
Therefore, the activation energy for reaction A to B is 100kJ
Multiple Choice:
1 e
2 c
3d
4c
5a
6d
7e
8b
9e